Example5: Find ∫ex sinx dx
نستخدم التكامل بالتجزيء
u = ex , dv = sinx dx Then
du = ex dx , v = – cosx
òu dv = u v –òv du
òex sinx dx = ex ( – cosx) – ò– cosx (ex) dx OR òex sinx dx = òex d/dx (– cosx) dx Then .... (1)
òex sinx dx =– ex cosx + òex cosx dx ..... (1)
Repeat it with New Integration òex cosx dx
u = ex , dv = cosx dx Then
du = ex dx , v = sinx
òu dv = u v – òv du
òex cosx dx = ex (sinx) – òsinx (ex) dx OR òex cosx dx = òex d/dx ( sinx) dx Then .... (2)
òex cosx dx = ex sinx – òex sinx dx ..... (2)
From (2) in (1)
òex sinx dx =– ex cosx + ex sinx – òex sinx dx
Shift from Left side to Right side
2òex sinx dx =– ex cosx + ex sinx ( / 2)
ex( sinx – cosx )
òex sinx dx = ——————— + c
2